Answer:
Time period of the osculation will be 0.0671 sec
Explanation:
It is given a vertical spring is stretched by 4 cm
So change in length of the spring x = 4 cm = 0.04 m
Mass which is hung from it m = 12 gram = 0.012 kg
Sprig force will be equal to weight of the mass
So 
k = 244.7 N/m
Now new mass is m = 28 gram = 0.028 kg
So time period with new mass will be
Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
Explanation:
Given that;
mass of vehicle m = 1000 kg
for a low speed test; V = 2.5 m/s
bumper maximum deflection = 4 cm = 0.04 m
First we determine the energy of the vehicle just prior to impact;
W_v = 1/2mv²
we substitute
W_v = 1/2 × 1000 × (2.5)²
W_v = 3125 J
now, the the effective design stiffness k will be:
at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;
hence;
W_v = 1/2kx²
we substitute
3125 = 1/2 × k (0.04)²
3125 = 0.0008k
k = 3125 / 0.0008
k = 3906250 N/m
Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m
Answer:
FB = 0.187 N
Explanation:
To find the magnetic force FB in the wire you use the following formula:
the angle between B and L is given by:
Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:
![|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N](https://tex.z-dn.net/?f=%7C%5Cvec%7BF_B%7D%7C%3DIsin%5Ctheta%5Cint_0%5E%7B0.25%7DB%28y%29dy%3DIsin%5Ctheta%5Cint_0%5E%7B0.25%7D%280.5y%29dy%5C%5C%5C%5C%7C%5Cvec%7BF_B%7D%7C%3D%282.0%2A10%5E%7B-3%7DA%29%28sin36.86%5C%C2%B0%29%280.5T%29%5B%5Cfrac%7B0.25%5E2%7D%7B2%7Dm%5D%3D0.187%5C%20N)
hence, the magnitude of the magnetic force is 0.187N
Explanation:
Always behave responsibly in the laboratory. Do not run around or play practical jokes. Always check the safety data of any chemicals you are going to use. Never smell, taste or touch chemicals unless instructed to do so.
Answer:
196000 N
Explanation:
The following data were obtained from the question:
Height (h) = 10 m
Area (A) = 2 m²
Force (F) =.?
Next, we shall determine the pressure in the tank.
This can be obtained as follow:
P = dgh
Where
P is the pressure.
d is the density of the liquid.
g is acceleration due to gravity
h is the height.
Height (h) = 10 m
Density (d) of water = 1000 kg/m³
Acceleration due to gravity (g) = 9.8 m/s²
Pressure (P) =...?
P = dgh
P = 1000 × 9.8 × 10
P = 98000 N/m²
Therefore, the pressure acting on the tank is 98000 N/m²
Finally, we shall determine the force of gravity acting on the column of water as follow:
Area (A) = 2 m²
Pressure (P) = 98000 N/m²
Force (F) =.?
Pressure (P) = Force (F) /Area (A)
P = F /A
98000 = F/ 2
Cross multiply
F = 98000 × 2
F = 196000 N
Therefore, the force of gravity acting on the column of water is 196000 N
